I have a finite measure space $(X, \mathcal{S}, \mu)$, and a transformation $f:X\rightarrow X$ that "preserves measure on average". That is, for $A \in \mathcal{S}$ $$ \lim_{n\rightarrow \infty} \frac{1}{n}\sum_{k=1}^n \mu(f^k(A)) = \mu(A) $$ I would like to be able to apply some ergodic or recurrence theorem in this situation, but Poincare Recurrence for example requires that the map $f$ be measure-preserving, which this "measure-preserving on average" mapping does not satisfy.

I am wondering if anyone can say if there are any results concerning such "measure-preserving on average" maps and any related recurrence or ergodic theorems, or if there is some way I can use say an altered Poincare Recurrence to makes claims about this mapping.

  • $\begingroup$ Your condition is a bit problematic because $f(A)$ may not be a measurable set. It's generally better to write invariance conditions in terms of $f^{-1}$. If a measure satisfies your condition with $f^{-1}$ in place of $f$, it actually is invariant: you can just substitute $f^{-1}(A)$ for $A$ and you see you have the same limit, so your condition implies $\mu(A)=\mu(f^{-1}A)$. $\endgroup$ – Anthony Quas Dec 7 '20 at 2:36

The phrase "a finite measure space $(X,\mathcal{S},\mu)$ where $\mu$ is the Lebesgue measure" can only reasonably mean that $\mu$ is a nonzero multiple of the counting measure, with $\mathcal{S}$ being the powerset of $X$. At least, we may assume that, if $A\subsetneq X$, then $\mu(A)<\mu(X)$. Therefore and because the set $X$ is finite, $$M:=\max\{\mu(A)\colon A\subsetneq X\}<\mu(X).$$ So, if $f(X)\ne X$, then your displayed condition implies $$\mu(X)\le\lim_{n\to\infty}\frac1n\,\sum_{k=1}^n M=M<\mu(X).$$ This contradiction implies that $f(X)=X$. Since the set $X$ is finite, this means that $f$ is a bijection or, in other words, permutation of the set $X$.

On the other hand, if every permutation of $X$ "preserves a measure $\mu$ on average", then it is easy to see that $\mu$ must be a multiple of the counting measure.

In particular, it follows that, if $\mu$ is a nonzero multiple of the counting measure on $X$, then a map $f\colon X\to X$ "preserves the measure $\mu$ on average" iff $f$ is a permutation of the set $X$.

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    $\begingroup$ It may be that "measure space with finite measure" is what was intended. Maybe OP can clarify… $\endgroup$ – Dirk Dec 6 '20 at 19:56
  • $\begingroup$ Yes I meant just measure space of finite measure, I am not sure why I added the Lebesgue measure part. $\endgroup$ – GEG Dec 6 '20 at 19:57
  • $\begingroup$ I edited the question appropriately, and if this changes much in your answer I would very much appreciate an explanation of what changes. $\endgroup$ – GEG Dec 6 '20 at 19:59
  • $\begingroup$ @GEG : I think "a finite measure space" can only mean a measure space which is finite, and that is something quite different from "a measure space with a finite measure" or "a finite-measure measure space". Of course, the answer entirely relies on the condition that the space is finite. $\endgroup$ – Iosif Pinelis Dec 6 '20 at 21:35

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